Re: Gaussian Curvature of a Sphere

1st coordinate, leave out "1" subscripts, 2-->3 is positive (the positive cycle goes like this: 1-->2-->3-->1-->2-->3 etc.) (so 3-->2 is negative) so the u2v3 term is positive.

2nd coordinate, leave out "2", the positive term is u3v1 (see above).

3rd coordinate, well, there's only one possible way to "balance it out" now.

i find this easier to remember than the "determinant formula":

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(by the way, you wrote down the wrong "third coordinate" in your parametrization, it should be: not theta. you did get the correct E,F and G, though)

a lengthy calculation shows that:

which is what you got (you can simplfy your "k" coordinate by taking out the common factor of ).

this is a normal vector, but it's not a UNIT vector, so we need to "normalize" it (how's that for an odd pun?) by dividing by its norm, which is:

this is almost as much work as finding the cross-product, first we find that:

taking the square root, we get that the magnitude of our normal vector is so that:

(a word about the signs, here: our parameterization takes horizontal rays vectors going left-to-right to "lattitude" arcs going counter-clockwise (west-to-east), and vertical rays going down-to-up to "longitude" arcs going from the north pole to the south pole, so the "outward" normal (using a right-hand-rule for the cross-product) in this case points "towards" the center of the sphere (inwards). we could have chosen a different parameterization to make the normal vector n simply be:

which would have been more intuitively obvious, geometrically (using -θ instead of θ, for example)).

i leave it to you to compute the second derivatives, and thus L,M and N.